Answer
The bubble's diameter is $2.10~mm$ when it reaches the surface.
Work Step by Step
We can find the gauge pressure at a depth of 80.0 meters:
$P_g = \rho~g~h = (1000~kg/m^3)(9.80~m/s^2)(80.0~m) = 7.84\times 10^5~Pa$
We can find the absolute pressure initially:
$P_1 = P_{atm}+P_g$
$P_1 = (1.01\times 10^5~Pa)+(7.84\times 10^5~Pa)$
$P_1 = 8.85 \times 10^5~Pa$
We can find an expression for the original volume:
$P_1~V_1 = nRT_1$
$V_1 = \frac{nRT_1}{P_1}$
We can find an expression for the final volume:
$P_2~V_2 = nRT_2$
$V_2 = \frac{nRT_2}{P_2}$
We can divide $V_2$ by $V_1$:
$\frac{V_2}{V_1} = \frac{\frac{nRT_2}{P_2}}{\frac{nRT_1}{P_1}}$
$V_2 = \frac{T_2~P_1}{T_1~P_2}~V_1$
$V_2 = \frac{(291~K)(8.85\times 10^5~Pa)}{(277~K)(1.01\times 10^5~Pa)}~V_1$
$V_2 = 9.205~V_1$
We can find the radius $r_2$ of the bubble at the surface:
$V_2 = 9.205~V_1$
$\frac{4}{3}\pi~r_2^3 = (9.205)~\frac{4}{3}~\pi~r_1^3$
$r_2^3 = (9.205)~r_1^3$
$r_2 = (9.205)^{1/3}~r_1$
$r_2 = (2.10)~(0.500~mm)$
$r_2 = 1.05~mm$
Since the diameter is twice the radius, the bubble's diameter is $2.10~mm$ when it reaches the surface.
